This question is pertaining to College Algebra

Question

The John Deere Company has found that the revenue from sales of heavy-duty trackers is a function of the unit price p that it charges. If the revenue R is R(p) = - 1/2p² + 1900p what unit price should be charged to maximize revenue? What is the maximum revenue?

Answer 1

It is parabola looking down and has the maximum, find the vertex of equation.(p,R)

vertex = -b/2a just substitute fro your equation

-1900/2(-1/2) = 1900 it is your x for vertex

put it in your equation and calculate:

R= -1/2(1900)^2 +1900x1900 = -1805000 + 3610000= 1805000

price must be equal 1900 to maximize the revenue = 1805000